My simple explanation is that the entropy never increased in the first place, and like a good magic trick, they showed you a state where it looked like mixing had occurred, but it in fact had not. In the false "mixed" state the colors were extremely systematically layered, looping around the apparatus several times.
Wikipedia actually does a far better job than I could with this, but I'll reference their equations anyway. Here is the typical Navier-Stokes equation beautifully written from the Wikipedia article. You find in the article about Stokes flow that in the case of a very low Reynolds number, the effective derivative part, the left hand side, is all equal to zero. Also, I think that $\mathbf{f}=-\rho g$ downward, but I'd like someone to confirm that fact.
$$\overbrace{\rho \Big(
\underbrace{\frac{\partial \mathbf{v}}{\partial t}}_{
\begin{smallmatrix}
\text{Unsteady}\\
\text{acceleration}
\end{smallmatrix}} +
\underbrace{\mathbf{v} \cdot \nabla \mathbf{v}}_{
\begin{smallmatrix}
\text{Convective} \\
\text{acceleration}
\end{smallmatrix}}\Big)}^{\text{Inertia (per volume)}} =
\overbrace{\underbrace{-\nabla p}_{
\begin{smallmatrix}
\text{Pressure} \\
\text{gradient}
\end{smallmatrix}} +
\underbrace{\mu \nabla^2 \mathbf{v}}_{\text{Viscosity}}}^{\text{Divergence of stress}} +
\underbrace{\mathbf{f}}_{
\begin{smallmatrix}
\text{Other} \\
\text{body} \\
\text{forces}
\end{smallmatrix}} = 0$$
In addition to this, the mass balance in an incompressible flow is the following.
$$\boldsymbol{\nabla}\cdot\mathbf{v}=0$$
The solution to these equations includes $\mathbf{v}$ (vector velocity) and $p$ (scalar pressure). The important part to understand is that the time derivative is gone. In this case, the second you stop putting torque into the machine, the fluid is stationary. So it's actually entirely the boundary conditions that define the movement of the fluid at any given time. I think it's fair to them make the following statement about the position and velocity.
$$\int_0^{t_1} \mathbf{v} dt = \mathbf{r}$$
This gives the "mapping" of one particle in the position to the position at the end of the experiment. The entire idea is that they can produce the conditions such that the following is also met.
$$\int_{t_1}^{t_2} \mathbf{v} dt = -\mathbf{r}$$
Ordinarily this would be the exact thing prevented by entropy. Basically, the equation leads to the creation of too much information to be exactly reversed by simple set of boundary conditions. I don't have any good way to write this, but the boundary conditions are sufficiently reversed in time and the rate at which it's turned also doesn't matter as a result of the form of the equations.
